Endgame #5 [ENDORSED]

$E_{cell} = E_{cell}^{\circ}-\frac{RT}{nF}\ln Q$

Posts: 50
Joined: Fri Sep 27, 2019 12:29 am

Endgame #5

In the Endgame review packet, #5 says:

"Calculate the value of Ka for HF using the standard cell potentials:

F2 + 2 H+ (aq) + 2e- --> 2 HF (aq) Eo = +3.03 V
F2 + 2 e- --> 2 F- (aq) Eo = +2.87 V "

I understand why Eo is -0.16 V but I don't understand why n (the moles of electrons transferred) is 1. In Lyndon's answer key, he doesn't explain why either. By adding the two half-reactions, he gets 2HF --> 2 H+ + 2 F- which he then reduces to HF --> H+ + F-, but aren't 2 moles of electrons still needed in order for this reaction to even proceed? Each half-reaction requires 2 moles of e- transferred in order to run therefore n should be 2.

If someone could explain this, I'd be incredibly thankful!

Jiyoon_Hwang_2I
Posts: 101
Joined: Sat Sep 14, 2019 12:17 am

Re: Endgame #5

I think n is 1 electron instead of 2 because the question asked for the Ka of HF and not 2HF

Chem_Mod
Posts: 18878
Joined: Thu Aug 04, 2011 1:53 pm
Has upvoted: 714 times

Re: Endgame #5  [ENDORSED]

Anuradha S 1F wrote:In the Endgame review packet, #5 says:

"Calculate the value of Ka for HF using the standard cell potentials:

F2 + 2 H+ (aq) + 2e- --> 2 HF (aq) Eo = +3.03 V
F2 + 2 e- --> 2 F- (aq) Eo = +2.87 V "

I understand why Eo is -0.16 V but I don't understand why n (the moles of electrons transferred) is 1. In Lyndon's answer key, he doesn't explain why either. By adding the two half-reactions, he gets 2HF --> 2 H+ + 2 F- which he then reduces to HF --> H+ + F-, but aren't 2 moles of electrons still needed in order for this reaction to even proceed? Each half-reaction requires 2 moles of e- transferred in order to run therefore n should be 2.

If someone could explain this, I'd be incredibly thankful!

You are correct that each half-reaction requires 2 moles of electrons to be transferred; that is how you obtain the overall reaction with coefficients of 2. It logically follows that halving the overall reaction will halve the moles of electrons transferred. It's the same as halving each half reaction and adding them together, which changes to 1 mole of electrons being transferred.

If you want to think of this another way, if you go into your chem 14A knowledge, to go from H-F to H+ and F-, the electron that H contributes to the covalent bond is transferred to F. This is a 1 electron transfer.

Return to “Appications of the Nernst Equation (e.g., Concentration Cells, Non-Standard Cell Potentials, Calculating Equilibrium Constants and pH)”

Who is online

Users browsing this forum: No registered users and 1 guest